Question

If $x=\cos t$ and $y=\ln t$ then the value of $\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2$ at $t=\frac{\pi }{2}$ is equal to

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (B)

$x=\cos t$
$\frac{dx}{dt}=-\sin t$
$y=\ln t$
$\frac{dy}{dt}=\frac{1}{t}$
$\frac{dy}{dx}=\frac{-1}{t\sin t}\Rightarrow {\left(\frac{dy}{dx}\right)}_{t=\pi /2}^2=\frac{4}{{\pi }^2}$
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{-1}{t\sin t}\right)=\frac{d}{dt}\left(\frac{-1}{t\sin t}\right)\frac{dt}{dx}$
$=\frac{1}{(t\sin t{)}^2}(t\cos t+\sin t)\left(\frac{-1}{\sin t}\right)$
$\frac{d^{2}y}{d{x}^2}=\frac{-t\cos t-\sin t}{t^2{\sin }^{3}t}$
${\frac{d^{2}y}{d{x}^2}|}_{t=\pi /2}=\frac{-4}{{\pi }^2}$
$\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0$ .